Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;

class Main {
    public static void main(String[] args) {
    	Scanner sc = new Scanner(System.in);

    	int n = sc.nextInt();

    	double[][] dp = new double[2][n+5];
    	int curr = 0;
    	int next = 1;
    	dp[curr][0] = 1;
    	for(int z = 0; z < n; z++) {
    		Arrays.fill(dp[next], 0);
    		double p = sc.nextDouble();
    		for(int i = 0; i < n; i++) {
    			dp[next][i+1] += dp[curr][i] * p;
    			dp[next][i] += dp[curr][i] * (1-p);
    		}
    		int tmp = curr;
    		curr = next;
    		next = tmp;
    	}

    	double ans = 0;
    	for(int i = (n+1)/2; i < n+1; i++)
			ans += dp[curr][i];

    	System.out.println(ans);
    }
}